ABC

The square of a real number is non-negative. Hence $a^{2}=b^{2}=c^{2}=0\implies (a,b,c)=\boxed{(0,0,0)}$ and we are done.

Square of a real number is always Non negative i.e. $(-1)^{2} = 1$ $(-2)^{2} = 4$ $(-0.5)^{2} = 0.25$ $(0)^{2} = 0$

so the sum of 3 non negative is equal to zero only when each number is equal to zero

$a^{2} = b^{2} = c^{2} = 0$ implies $(a,b,c) = (0,0,0)$

Something is wrong, they said a,b and c are real numbers [only positive], so how come they all can be zero ?

Square of a real number is always positive. Thus, if 3 positive integers add up to zero there is only one possibility that all the numbers are zeroes.

It's not defined because a,b,c are different variables, they have different values. They can't be 0,0,0

a=0 b=0 c=0 0^2=0 zero squre values of also calculate zero ( a b c)=(0 0 0)

the answer is 0 0 0

It can be also that at least one of the a or b or c is a complex number. This will give a negative value that will get the 0.

if we square any positive or negative number, we will get a positive answer.For example, (1)^2=1 and (-1)^2=1.So, we can never get a negative answer by squaring something. The only way to satisfy the equation is when all a,b,c are equal to zero.,

since 0 square is equivalent to 0, hence all variables should be zero and not 1 whose square is still 1

a^2+b^2+c^2=0 ,a=b=c=0

0squ.+0squ.+osqu=0 n ans 0,0,0

$(0,0,0)\quad is\quad right\quad answer\\ Because\\ { 0 }^{ 2 }{ +0 }^{ 2 }{ +0 }^{ 2 }=0\\ others\quad are\quad incorrect$

It is very easy ; since , a.a + b.b + c.c =0 { says equation 1 } by equation 1 we have , a.a = b.b = c.c = 0

so ; ( a,b,c ) = ( 0,0,0 ) .

since sqare of real numbers ae always non-negative,if sum of three non-negative numbers are 0 then they must be 0.

in all option only this option is having real number so ans is 0,0,0

-1 square will give you +1 so only one possibility i.e. 0 0 0